Math 431 Homework 6
Due 10/30
1. Let P and Q be two points and l and m two lines. What can you say
about these points and lines if you know that side(P, l) ∩ side(Q, m) = ∅? In
the event that there is a point L ∈ {l} such that P ∗ L ∗ Q and M ∈ {m}
such that P ∗ M ∗ Q show that L ∗ M ∗ Q and P ∗ L ∗ M .
Solution: We have side(P, l)∩ side(Q, m) = ∅. We first note that P ∈
/ {l}
and Q ∈
/ {m}. There are three possibilities:
1. l and m are distinct lines that share a point,
2. l and m are parallel, and
3. l = m
Let us consider each case in turn.
1. l and m are distinct lines that share a point. Then either
(a) P lies on m
i. Q lies on l: Let R be a point such that P ∗ R ∗ Q, which
exists by axiom B-2. By Lemma 3.2.2 P and R are on the
same side of l and R and Q are on the same side of m. Hence,
P ∈side(P, l) ∩ side(Q, m), which is a contradiction.
ii. Q does not lie on l. First note that P and Q have to be on
opposite sides of l, for if they were not then Q would belong
to side(P, l) ∩ side(Q, m), which contradicts the assumption.
Since P and Q are on opposite sides of l the segment P Q
intersects l in a point, call it L, so that P ∗ L ∗ Q. By axiom
B-2 there is a point R such that L ∗ R ∗ P . Using Proposition
3.3, we can conclude that Q ∗ R ∗ P , that is Q and P are on
the same side of m. Further, since L ∗ R ∗P we see that P and
R are on the same side of l, hence R ∈side(P, l) ∩ side(Q, m),
contradicting our assumption.
(b) P does not lie on m
i. Q lies on l: proof follows the proof in case of P ∈ {m}, Q ∈
/
{l}.
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ii. Q does not lie on l. Take any point M on m so that P and
M are on the same side of l (such point exist, because if it
did not then m and l would be parallel, which contradicts
our hypothesis), and take any point L on l so that Q and L
are on the same side of m. By axiom B-2 there is a point R
such that L ∗ R ∗ M . By Lemma 3.2.2 and axiom B-4 R and
P are on the same side of l. Similarly, Q and R are on the
same side of m, hence R ∈side(P, l) ∩ side(Q, m), once again
contradicting our assumption.
Hence, this case can not happen, given our hypothesis.
2. l||m. As in the previous case we have few possible configurations depending on where the given points lie with respect to the given lines.
(a) P lies on m
i. Q lies on l. Let R be a point such that P ∗ R ∗ Q. Using
Lemma 3.2.2 we conclude that R and P are on the same
side of l and R and Q are on the same side of m, that is
R ∈side(P, l) ∩ side(Q, m).
ii. Q does not lie on l. As we noted above P and Q must lie on
opposite sides of l, so by Lemma 3.2.5 there is a point L ∈ {l}
such that P ∗ L ∗ Q. Let R be such that P ∗ R ∗ L (B-1). Then
Lemma 3.2.2 and B-4 give us that R ∈ side(P, l). Similarly,
R ∈ side(Q, m). Contradiction.
(b) P does not lie on m
i. Q lies on l. As 2(ii).
ii. Q does not lie on l. Using the arguments above we can show
that P and Q must lie on opposite sides of both l and m. Let
L ∈ {l} such that P ∗L∗Q and M ∈ {m} such that P ∗M ∗Q.
By B-3 one of the following happens:
• P ∗ M ∗ L – then P and M are on the same side of l
(Lemma 3.2.5). Let S be such that M ∗ S ∗ L, so that M
and S are on the same side of l. By B-4, P and S are on
the same side of l. Also, P ∗ M ∗ L and P ∗ L ∗ Q give us
M ∗ L ∗ Q bu Proposition 3.3, so L and Q are on the same
side of m (Lemma 3.2.2). From M ∗ S ∗ L we conclude
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that L and S are on the same side of m, so by B-4 we
have that S and Q are on the same side of m. Hence,
S ∈side(P, l) ∩ side(Q, m).
• P ∗ L ∗ M together with P ∗ M ∗ Q gives, by Proposition
3.3 L ∗ M ∗ Q, hence our claim holds.
• M ∗ P ∗ L together with P ∗ L ∗ Q gives M ∗ P ∗ Q, so P
and Q are on the same side of m, contradiciton.
Note: I could have done this whole argument without using
Proposition 3.3, but that would have made it much longer
than it already is, so I chose not to.
3. l = m. P and Q lie on opposite sides of l, for if they did not then
side(P, l) = side(Q, m), so side(P, l) is their intersection, and that set
is nonempty.
This exhaust all the possible cases.
2. Prove Proposition 3.8: If D is in the interior of an CAB then:
−−→
1. so is every point on AD except A,
−−→
2. no point on the opposite ray to AD is in the interior of BAC
3. if C ∗ A ∗ E, then B is in the interior of DAE
−−→
Proof of (1). Suppose that E ∈ AD and E 6= A. Since D is in the interior
←→
of CAB, D and B are on the same side of AC; by Lemma 3.7.5, E and D
←→
are on the same side of AC; hence, by B-4 E and B are on the same side
←→
←→
of AC. By the same reasoning, since D and C are on the same side of AB,
we deduce from Lemma 3.7.5 and B-4 that E and C are on the same side of
←→
AB. Thus E is in intCAB.
−−→
Proof of (2). Suppose that E is on the ray opposite to AD. Then E = A or
−−→
E ∗ A ∗ D by definition of the ray opposite to AD (as discussed in class). If
←→
←→
E = A, then E lies on AC, so E and B are not on the same side of AC, so
E is not in the interior of CAB. Suppose that E ∗ A ∗ D. By Lemma 3.2.4,
←→
E and D are on opposite sides of AB. Since D ∈ intCAB, D and C are
←→
on the same side of AB. Thus, by Corollary to B-4 E and C are on opposite
←→
sides of AB. Thus, E is not in intCAB.
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Proof of (3). There are two things to show: first is B and D are on the same
←→
←→
side of AE, and second is B and E are on the same side of line AD.
←→ ←→
Since C ∗ A ∗ E, by B-1 and I-1 we have AC = AE. Since D is in the
←→
←→
interior of CAB, B and D are on the same side of AC, hence AE.
←→
To prove that B and E are on the same side of AD, suppose on the
←→
contrary that E and B not on the same side of AD. Before we can say that
←→
B and E are on opposite sides of AD, we must first check that neither B nor
←→
←→
E is on AD. Since D ∈ intCAB, D and C are on the same side of AB, so
←→
D is not on AB (definition of same sides), so A, B, D are not collinear, so B
←→
is not on AD. Also, since D ∈ intCAB, D and B are on the same side of
←→ ←→
←→
AC = AE, so D is not on AE, so A, D, E are not collinear, so E is not on
←→
←→
AD. Now that we know that B and E do not lie on AD and are not on the
←→
←→
same side of AD, they must be on opposite sides of AD.
By definition of opposite sides and segment, there is a point F lying on
←→
AD such that E ∗ F ∗ B. By Proposition 3.7, F is in the interior of BAE.
←→
(Note that BAE is an angle because B does not lie on AE.) In particular
←→
F and B are on the same side of AE. Since B and D are on the same side
←→
←→
←→
of AC = AE, by B-4 F and D are on the same side of AE. Thus, either
D = F , A ∗ D ∗ F or A ∗ F ∗ D by Lemma 3.2.3. In any case, D is on ray
−→
−→
AF . Thus, by Proposition 3.8(a) (applied to AF and BAE) D is in the
interior of BAE.
←→
By definition of interior, D and E are on the same side of AB. We also
←→
know that D and C are on the same side of AB because D is in interior of
←→
CAB. Therefore, by B-4(i), C and E are on same side of AB. But, since
←→
C ∗ A ∗ E, by Lemma 3.2.4 C and E are on opposite sides of AB. This is a
←→
contradiction. Therefore, B and E are on the same side of AD.
3. If B and D are distinct points there exists a point C such that B∗C∗D.
←→
1. There exists line BD through B and D – by axiom I-1, since B and D
are distinct points.
←→
2. There exists a point F not lying on BD – by Proposition 2.3.
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←→
3. There exists a line BF through B and F – by axiom I-1, since F does
←→
←→ ←→
not lie on BD we must have F 6= B, and also BD 6= BF .
4. There exists a point G such that B ∗ F ∗ G – by axiom B-2, since B
and F are distinct points.
5. Points B, F, and G are collinear – by axiom B-1 and step 4.
6. G and D are distinct points and D, B and G are not collinear – G
←→
←→
lies on BF , D lies on BD, and the intersection of those two lines is
B. Since the lines are distinct (step 3), by Proposition 2.1 B is the
only point they have in common, hence G and D are distinct points.
If D, B and G were collinear, they would have to lie on a unique line
←→
←→ ←→
BD (axiom I-1), so BD = BF which contradicts step 3.
7. There exists a point H such that G ∗ D ∗ H – step 6 guarantees that
we can apply axiom B-2 to points G and D.
←→
8. There exists a line GH – by axiom I-1.
9. H and F are distinct points – If they were the same then we would
have B ∗ F ∗ G and G ∗ D ∗ F , so by axiom B-1 B, G and D are collinear
points contradicting step 6.
←→
10. There exists a line F H – previous step and axiom I-1.
←→
←→
←→ ←→
11. D does not lie on F H – F does not lie on GD (step 9), so GD 6= F H.
Since H lies on each of those lines, and since H 6= D by step 7, by
Proposition 2.1, D does not.
←→
12. B does not lie on F H – If it did, then H would lie on the unique line
←→
←→
←→
BF determined by B and F (axiom I-1). Lines BF and GD now have
two points in common: G and H. By Proposition 2.1 they would have
to be equal, which contradicts the previous step.
←→
13. G does not lie on F H – if it did we would have: G, F, H collinear,
G, D, H collinear, hence G, D, B collinear (usinga axiom I-1) which
contradicts step 6.
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14. Points D, B and G determine 4DBG – step 6 and definition of a triangle,
←→
and F H intersects side BG in a point between B and G – steps 4, 12
and 13.
←→
←→
15. H is the only point lying on both F H and GH – these two lines are
distinct, eg. step 13, so by Proposition 2.1 they share exactly one point:
H.
←→
16. No point between G and D lies on F H – step 7 and axiom B-3.
←→
17. Hence, F H intersects side BD in a point C between D and B – step 14,
16 and Pasch’s theorem (note that it can’t be point D since G ∗ D ∗ H).
18. Thus, there exists a point C between points B and D.
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